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katen-ka-za [31]
2 years ago
5

Helppppp mathhhhhhhhhhhh

Mathematics
1 answer:
antoniya [11.8K]2 years ago
3 0
The first one is the correct answer, enjoy!
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What is the measure of AngleJHG?
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There is no attachment so we don’t know what this angle is.
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What is the difference between the greatest green bean weight and the least green bean weight.
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The Bigger weight minus the lower weight
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3 years ago
Find three consecutive odd integers such that the sum of the 1st and 2nd number is 344
olga2289 [7]

Answer:

three consecutive odd integers are.

171, 173, 175

Step-by-step explanation:

Let x be the first term of odd integer,

Therefore, the second consecutive odd integers = x+2

Similarly, the third consecutive odd integers = x+4

Given:

The sum of the first term and second term is 344.

first term + second term = 344

Here first term is x and second term is x+2.

x+(x+2)=344

2x+2=344

2x=344-2

x=\frac{342}{2}

x=171

Therefore, the first odd integers is 171.

And the second consecutive odd integers = x+2=171+2=173

Similarly, the third consecutive odd integers = x+4=171+4=175

8 0
3 years ago
Which function is undefined for x = 0? y=3√x-2 y=√x-2 y=3√x+2 y=√x=2
Andre45 [30]

For this case, we must indicate which of the given functions is not defined forx = 0

By definition, we know that:

f (x) = \sqrt {x} has a domain from 0 to infinity.

Adding or removing numbers to the variable within the root implies a translation of the function vertically or horizontally. For it to be defined, the term within the root must be positive.

Thus, we observe that:

y = \sqrt {x-2} is not defined, the term inside the root is negative when x = 0.

While y = \sqrt {x + 2} if it is defined for x = 0.

f(x)=\sqrt[3]{x}, your domain is given by all real numbers.

Adding or removing numbers to the variable within the root implies a translation of the function vertically or horizontally. In the same way, its domain will be given by the real numbers, independently of the sign of the term inside the root.

So, we have:

y = \sqrt [3] {x-2} with x = 0: y = \sqrt [3] {- 2} is defined.

y = \sqrt [3] {x + 2}with x = 0: y = \sqrt [3] {2}in the same way is defined.

Answer:

y = \sqrt {x-2}

Option b


6 0
3 years ago
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